tui tk cho ai đúng nhe

\(a,1,3\times0,5< x< 1,8\times1,9\\ \Leftrightarrow0,65< x< 3,42\\ \Rightarrow x=1,x=2,x=3\)

\(b,2,5\times2,7< x< 2,7\times4,5\\ \Rightarrow6,75< x< 12,15\\ \Rightarrow x=7;x=8;x=9;x=10;x=11;x=12\)

khó đấy

huhu

a) 0

b) 0

Khó đấy !!!

😵😵😵😵

a=68  vì ucln(680,884)=68

\(VT\ge0\Rightarrow\)\(-5\le y\le5\)

\(VT=8k^2\Rightarrow25-y^2=8k^2\Rightarrow k^2\le3\)
\(k^2=\left\{0,1\right\}\)

\(k=0\Rightarrow\hept{\begin{cases}x=2009\\y=+-5\end{cases}}\)

\(k^2=1\Rightarrow y^2=17\left(loai\right)\)

KL

\(\left(x,y\right)=\left(2009,-5\right);\left(2009,5\right)\)

a=6;b=4;c=6

Số tự nhiên A là :

16 - 8 = 8

Số tự nhiên B là :

8 - 4 = 4

Số A gấp số B hay số tự nhiên C là :

8 : 4 = 2

X x 2011-X=2011x2009+2011

Xx(2011-1)=2011x(2009+1)

Xx2010    = 2011x2010

=> X=2011

x=2011

x * 2011 - x = 2011 x 2009 + 2011

x * ( 2011 - 1 ) = 2011 x ( 2009 + 1 )

x * 2010 = 2011 x 2010

x = 2011

a, x ⋮ 25 và x < 100

Vì x ⋮ 25 

nên x ∈ B(25) = { 0;25;50;75;100;... }

Mà x < 100

=> x = { 0 ; 25 ; 50 ; 75 }

 b,5x + 3x = 3^6 : 3^3 .4 + 12

   x.( 5 +3 )= 3^3 . 4 + 12

    x . 8       = 27 . 4 + 12

    x . 8       = 108 + 12

    x . 8       = 120

    x            = 120 : 8

    x            = 15

                                                               ~HT~

\(25-y^2-8.\left(x-2009\right)^2\)

ta thấy vế phải \(8.\left(x-2009\right)^2\ge0\) \(\forall x\)

\(\Rightarrow VT:25-y^2\ge0\)

\(\Rightarrow0\le y^2\le25\)

\(\Rightarrow y^2\in\left\{0;1;4;9;16;25\right\}\)

mà \(8.\left(x-2009\right)^2\) chẵn\(\Rightarrow25-y^2\)chẵn \(\Rightarrow y^2lẻ\)

\(\Rightarrow y^2\in\left\{1;9;25\right\}\)

\(\Rightarrow y\in\left\{1;3;5\right\}\) (do \(y\in N\))

\(TH1:y=1\)

\(\Rightarrow8.\left(x-2009\right)^2=24\)

\(\Leftrightarrow\left(x-2009\right)^2=3\left(koTM\right)\)(do \(x\in N\))

\(TH2:y=3\)

\(\Rightarrow8.\left(x-2009\right)^2=16\)

\(\left(x-2009\right)^2=2\left(koTM\right)\)(do \(x\in N\))

\(TH3:y=25\)

\(\Rightarrow8.\left(x-2009\right)^2=0\)

\(\Rightarrow\left(x-2009\right)^2=0\Rightarrow x=2009\left(TM\right)\)

vậy cặp số \(\left(x,y\right)\) thỏa mãn \(25-y^2-8.\left(x-2009\right)^2\)  là  \(\left(2009;25\right)\)

Đặt vế trái là A ta có:

\(\frac{A}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\)

\(\frac{A}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\)

\(\frac{A}{2}=\frac{1}{2}-\frac{1}{x+1}\Rightarrow\frac{A}{2}=\frac{x+1-2}{2\left(x+1\right)}\Rightarrow A=\frac{x-1}{x+1}\)

\(\Rightarrow\frac{x-1}{x+1}=\frac{2007}{2009}\Leftrightarrow x=2003\)
 

​\(\frac{A}{2}=\frac{1}{2}-\frac{1}{x+1}\Rightarrow\frac{A}{2}=\frac{x+1-2}{2\left(x+1\right)}\Rightarrow...

Đặt vế trái là A ta có:

\frac{A}{2}=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}2A​=61​+121​+201​+...+x(x+1)1​=2.31​+3.41​+4.51​+...+x(x+1)1​

\frac{A}{2}=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{x+1-x}{x\left(x+1\right)}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}2A​=2.33−2​+3.44−3​+4.55−4​+...+x(x+1)x+1−x​=21​−31​+31​−41​+41​−51​+...+x1​−x+11​